What is the general solution for sin^2(x)-7cos(x)-7=0 ?

The general solution for sin^2(x)-7cos(x)-7=0 is x=pi+2pin

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sin^2(x)-7cos(x)-7=0

Solution

sin^{2} (x) - 7 cos (x) - 7 = 0

Solution

x = π + 2 πn

Degrees

x = 18 0^{\circ} + 36 0^{\circ} n

Solution steps

sin^{2} (x) - 7 cos (x) - 7 = 0

Rewrite using trig identities

- 7 + sin^{2} (x) - 7 cos (x)

Use the Pythagorean identity:

cos^{2} (x) + sin^{2} (x) = 1

sin^{2} (x) = 1 - cos^{2} (x)

= - 7 + 1 - cos^{2} (x) - 7 cos (x)

Simplify

= - cos^{2} (x) - 7 cos (x) - 6

- 6 - cos^{2} (x) - 7 cos (x) = 0

Solve by substitution

- 6 - cos^{2} (x) - 7 cos (x) = 0

Let:

cos (x) = u

- 6 - u^{2} - 7 u = 0

- 6 - u^{2} - 7 u = 0 : u = - 6, u = - 1

- 6 - u^{2} - 7 u = 0

Write in the standard form

a x^{2} + b x + c = 0

- u^{2} - 7 u - 6 = 0

Solve with the quadratic formula

- u^{2} - 7 u - 6 = 0

Quadratic Equation Formula:

For

a = - 1, b = - 7, c = - 6

u_{1, 2} = \frac{- ( - 7 ) \pm ( - 7 ) ^{2} - 4 ( - 1 ) ( - 6 )}{2 ( - 1 )}

u_{1, 2} = \frac{- ( - 7 ) \pm ( - 7 ) ^{2} - 4 ( - 1 ) ( - 6 )}{2 ( - 1 )}

(- 7)^{2} - 4 (- 1) (- 6) = 5

(- 7)^{2} - 4 (- 1) (- 6)

Apply rule

- (- a) = a

= (- 7)^{2} - 4 \cdot 1 \cdot 6

Apply exponent rule:

(- a)^{n} = a^{n},

n

is even

(- 7)^{2} = 7^{2}

= 7^{2} - 4 \cdot 1 \cdot 6

Multiply the numbers:

4 \cdot 1 \cdot 6 = 24

= 7^{2} - 24

7^{2} = 49

= 49 - 24

Subtract the numbers:

49 - 24 = 25

= 25

Factor the number:

25 = 5^{2}

= 5^{2}

Apply radical rule:

5^{2} = 5

= 5

u_{1, 2} = \frac{- ( - 7 ) \pm 5}{2 ( - 1 )}

Separate the solutions

u_{1} = \frac{- ( - 7 ) + 5}{2 ( - 1 )}, u_{2} = \frac{- ( - 7 ) - 5}{2 ( - 1 )}

u = \frac{- ( - 7 ) + 5}{2 ( - 1 )} : - 6

\frac{- ( - 7 ) + 5}{2 ( - 1 )}

Remove parentheses:

(- a) = - a, - (- a) = a

= \frac{7 + 5}{- 2 \cdot 1}

Add the numbers:

7 + 5 = 12

= \frac{12}{- 2 \cdot 1}

Multiply the numbers:

2 \cdot 1 = 2

= \frac{12}{- 2}

Apply the fraction rule:

\frac{a}{- b} = - \frac{a}{b}

= - \frac{12}{2}

Divide the numbers:

\frac{12}{2} = 6

= - 6

u = \frac{- ( - 7 ) - 5}{2 ( - 1 )} : - 1

\frac{- ( - 7 ) - 5}{2 ( - 1 )}

Remove parentheses:

(- a) = - a, - (- a) = a

= \frac{7 - 5}{- 2 \cdot 1}

Subtract the numbers:

7 - 5 = 2

= \frac{2}{- 2 \cdot 1}

Multiply the numbers:

2 \cdot 1 = 2

= \frac{2}{- 2}

Apply the fraction rule:

\frac{a}{- b} = - \frac{a}{b}

= - \frac{2}{2}

Apply rule

\frac{a}{a} = 1

= - 1

The solutions to the quadratic equation are:

u = - 6, u = - 1

Substitute back

u = cos (x)

cos (x) = - 6, cos (x) = - 1

cos (x) = - 6, cos (x) = - 1

cos (x) = - 6 :

No Solution

cos (x) = - 6

- 1 \leq cos (x) \leq 1

No Solution

cos (x) = - 1 : x = π + 2 πn

cos (x) = - 1

General solutions for

cos (x) = - 1

cos (x)

periodicity table with

2 πn

cycle:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

x = π + 2 πn

x = π + 2 πn

Combine all the solutions

x = π + 2 πn

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Frequently Asked Questions (FAQ)

What is the general solution for sin^2(x)-7cos(x)-7=0 ?
The general solution for sin^2(x)-7cos(x)-7=0 is x=pi+2pin